Opposite sign correlations between variance components

Hi

It would be easier to address your question if I could see the results. However, it does happen that some variance components covary positively while others covary negatively. This situation arises with appreciable frequency, which is why the notion of "bivariate heritability" is fundamentally flawed. Suppose A contributes .1 to covariance, C contributes -.2 and E .3. Here the phenotypic covariance is .2, the simple sum of the effects. Trying to convert the contributions to proportions (which is what a heritability is) is nonsensical when there are positive and negative elements. Here cA/(cA+cC+cE) where c is contribution, yields .5 which seems ok, but then cC / (cA+cC+cE) = -1 and cE / (cA+cC+cE) = 1.5. There is no way that these should be regarded as proportions. The somewhat reasonable .5 for cA's contribution is incorrect. At the end of the day, I recommend looking directly at the contributions to covariance without trying to convert them to proportions. Taking absolute values of cA cC and cE and at the end of the calculations reimposing the sign can also give misleading impressions. So IMO, let's stick with answering the question how much does each variance component contribute to the covariance.

Hi

It would be easier to address your question if I could see the results. However, it does happen that some variance components covary positively while others covary negatively. This situation arises with appreciable frequency, which is why the notion of "bivariate heritability" is fundamentally flawed. Suppose A contributes .1 to covariance, C contributes -.2 and E .3. Here the phenotypic covariance is .2, the simple sum of the effects. Trying to convert the contributions to proportions (which is what a heritability is) is nonsensical when there are positive and negative elements. Here cA/(cA+cC+cE) where c is contribution, yields .5 which seems ok, but then cC / (cA+cC+cE) = -1 and cE / (cA+cC+cE) = 1.5. There is no way that these should be regarded as proportions. The somewhat reasonable .5 for cA's contribution is incorrect. At the end of the day, I recommend looking directly at the contributions to covariance without trying to convert them to proportions. Taking absolute values of cA cC and cE and at the end of the calculations reimposing the sign can also give misleading impressions. So IMO, let's stick with answering the question how much does each variance component contribute to the covariance.

HTH, Mike